Re: [avr-libc-dev] [bugs #12040] sbi in FAQ

[Erik Walthinsen]

John Altstadt wrote:
> Out of curiosity on my part, why do you think that
>    PORTB |= 0x03;
> and
>    PORTB |= 0x01;
>    PORTB |= 0x02;
> are equivalent statements?

They either are or aren't depending on what register you're 
manipulating.  You are totally correct, in certain registers or in 
hardware where PORTB timings are critical, the effects would different.

However, in the |= 0x03 case, the compiler read the register first, 
modifies it, then writes it out.  That can also cause improper operation 
and side effects depending on the register being accessed.

So the question is can the compiler optimize the case where two bits are 
set or cleared by using sbi/cbi, without stomping on certain usages?  I 
can see that either method can cause problems, the question is which one 
is more potentially harmful vs. the instruction savings?

The read/modify/write method takes longer and thus could cause problems 
especially in cases like the WDTCR's 4-cycle window.  This can be 
avoided by explicitely using the two-operation method.

The dual-sbi/cbi method is a cycle shorter and the effect is essentially 
immediate.  However, there would be no obvious way to revert to the 
r/m/w method, which could be bad.  OTOH, the dual-sbi/cbi method only 
functions for the lower 32 registers, which in the few AVR devices I've 
checked are pretty much UART, TWI, and PORT registers.  Not using r/m/w 
for two-bit operations on such registers may or may not be an issue, 
would need to think through more scenarios.

Comments?